Czech and Slovak Mathematical Olympiad

Consider linear functions $f(x) = a x + b$, $g(x) = b x + a$. Since $a, b$ are distinct and positive, their graphs are two distinct lines with positive slope. As $f(1) = g(1) = a + b$, point $P = [1, a + b]$ is the intersection of these lines (Fig. 3).

Without loss of generality, assume $b > a$ (i.e. the line determined by $g$ is the “steeper” one). Then $f(x) > g(x)$ for $x < 1$, whereas $f(x) < g(x)$ for $x > 1$: indeed,
$$f(x) - g(x) = (a x + b) - (b x + a) = (b - a)(1 - x).$$

Fig. 3

Let $t = \lfloor a + b \rfloor$ and consider $x_1 \le 1 < x_2$ such that $g(x_1) = t$ and $g(x_2) = t + 1$ (that is, $x_1 = \frac{t - a}{b}$ and $x_2 = \frac{t + 1 - a}{b}$). We claim that the interval $[x_1, x_2]$ has all the desired properties.

First, for any $x \in [x_1, x_2)$ we have
$$t = g(x_1) \le \min\{f(x), g(x)\} \le \max\{f(x), g(x)\} < g(x_2) = t + 1,$$
and thus $x$ is a solution to the equation.

Second,
$$1 = (t + 1) - t = b x_2 + a - (b x_1 + a) = b(x_2 - x_1),$$
and thus $x_2 - x_1 = 1 / b = 1 / \max\{a, b\}$ and the interval has the desired length.